Capacitor

IIT- PHY - CP 1

CAPACITOR

CAPACITOR

Capacitance, Parallel Plate capacitor with and without dielectric, capacitors in series and parallel, Energy stored in a capacitor.

POSITION VECTOR :

Capacitor is a device for storing electric charge and energy. It consists of a pair of conductors carrying equal and opposite charges (generally). Magnitude of this charge is known as the charge on the capacitor. Potential difference (V) between the two conductors is proportional to the charge on the capacitor (Q).

Q

α

V; Q = CV

Here the proportionality constant C is known as the capacitance of the capacitor. Capacitance depends on the size and shape of the plates and the material between them. The SI unit of capacitance is farad (F).

PARALLEL PLATE CAPACITORS :

A parallel plate capacitor consists of two equal flat parallel metal plates facing each other and separated by a dielectric of electric permittivity ∈.The plates may be square, rectangular or circular in shape.

For calculating the capacity of a capacitor, we first calculate electric field at a point between the plates and then using relation 

     − = dr dV

E compute the potential difference between the plates. Finally dividing the magnitude of charge (given to one plate) by the potential difference between the plates, we get the capacity.

In case of parallel plate capacitor as shown in figure. The field at P

(

)

E 2 2

σ

σ

σ

ε

ε

ε

− = − = or, ε σ = − dx dV [ as dx dV E=− ] or,

∫

∫

ε σ = − d 0 0 V dx dV i.e., V d ε σ = So,

₍

₎

d A / d A V q C =ε ε σ σ = = or

d

KA

C

=

ε

0 [as

ε ε

=

0

K

]

SPHERICAL CAPACITOR :

We can derive the capacity of a spherical capacitor in a similar

S Y L L A B U S

ε

+

+

+

+

+

+

+

+

−

−

−

−

−

−

−

−

0

=

x

x

=

d

V 0

=

V

P

•

σ

+

d

−

σ

q + + b a E + + +

−

−

−

−

−

−

−

IIT- PHY - CP 2 way V= 0 4 q πε _ −b_ 1 a 1 ; C = V q ₌

b

1

a

1

4

₀

−

πε

=

a

b

ab

4

₀

−

πε

If the radius of the outer sphere tends to infinity, b→

∞

, the capacitance reduces to C = 4πε0a which is the capacitance of the isolated sphere.

CYLINDRICAL CAPACITOR :

E = r 2πε0 λ a≤r≤b V = V_a −V_b = -

∫

a bEdr or V = 2πε₀ λ ln a b ∴ C = V q = V  λ ₌ a b n 2 ₀   πε ILLUSTRATION : 01

A parallel plate capacitor has plates of area 200cm2 _{and separation between the}

plates 1.00mm. What potential difference will be developed if a charge of 1.00nC (i.e., 1.00 x 10-9_{C) is given to it. Now if separation between the plates is increased to 2.00mm, what}

will be the new potential difference?

SOLUTION : The capacitance of the capacitor is

d

A

C

=

ε

0 = 8. 85 x 10-12 1 x 10 m m 10 x 200 x m F 3 -2 -4 = 0.177 x 10-9_F = 0.177nF.

The potential difference between the plates is

65 . 5 nF 177 . 0 nC 1 C Q V= = = volts.

If the separation is increased from 1.00mm to 2.00mm the capacitance is decreases by a factor of 2. If the charge remains the same, the potential difference will increase by a factor of 2. Thus, the new potential difference will be 5.65volts x 2 = 11.3 volts.

COMBINATION OF CAPACITORS

SERIES COMBINATION :

₊

_Q

₋

_Q

₊

_Q

₋

_Q

₊

_Q

₋

_Q

l

Q

IIT- PHY - CP

Capacitors connected as shown in the figure are said to be connected in series. In series combination the charges on individual condensers are equal and the total p.d. across the combination is to shared by the capacitors.

Q = C₁V₁=C₂V₂=C₃V₃ and V = V₁+V₂ +V₃

∴ Effective capacitance of the combination C can be found from the relation.

C 1 ₌ 1 C 1 + 2 C 1 + 3 C 1

PARALLEL COMBINATION :

In this combination p.d. across each of capacitors is same but the charge supplied at points A and B is shared by capacitors.

V = 1 1

C

Q

= 2 2

C

Q

= 3 3

C

Q

and total charge Q = Q₁ + Q₂ + Q₃ C = C₁ + C₂ + C₃ ILLUSTRATION : 02

Find the equivalent capacitance between points A and B of the circuit shown, each capacitance = C

SOLUTION : Equivalent circuit is

Each branch equivalent capacitance is

2 C

. There are four branches in parallel

∴Ceq = 2 C + 2 C + 2 C + 2 C = 2C ILLUSTRATION : 03

The figure shown is a system of parallel conductors. Each plate is of equal area A and equally separated by d. Find the equivalent capacitance of the system between a and

b

SOLUTION : By joining the points of same potential, the

arrangement of conductors may be reduced as shown in figure. If the capacitance between two successive plates is given by

d

A

C

=

ε

0 then, the equivalent capacitance of the

system is given by

d

A

2

3

2

C

3

C

_eq

=

=

ε

0 1

Q

+

−

-

Q

1 2

Q

+

-

Q

₂

−

3

Q

+

-

Q

₃

−

A

B

A B

A

B

a b 2 2 1 4 3 3

IIT- PHY - CP

DIELECTRICS :

When a dielectric is introduced between conductors of a capacitor, its capacitance increases. A dielectric is characterized by a constant 'K' called dielectric constant.

DIELECTRIC CONSTANT :

When a dielectric is placed in an external electric field, polarization occurs and it develops an electric field in opposition to the external one. As a result total field inside it decreases. If E is the total field inside the dielectric when it is placed in an external field

0

E , then its dielectric constant 'K' is given as K =

E

E

₀

( k > 1)

If a dielectric completely occupies the space between the conductors of a capacitor its capacitance increases 'K' times. Hence in presence of a dielectric with dielectric constant ' K ', the capacitance of a parallel plate capacitor =

d

A

K

ε

₀

ENERGY STORED IN A CAPACITOR :

The energy stored in a capacitor is equal to the work done to charge it. Let q be the instantaneous charge on either plate of the capacitor and the potential difference between the plate is V=

C q

. The work done to transfer an infinitesimal charge dq from the negative

plate to the positive plate is dW = Vdq =       C q dq

[The charge moves through the wires, not across the gap between the plates] ∴ W = total work done to transfer charge Q =

∫

Q 0 C q dq = C 2 Q2 ₌ 2 QV ₌ 2 1 _{C 2} V

This work done is stored as electrostatic energy ie., U = 2 1 C_V2₌ 2 1

_









 ε

d

A

0

(

_E2_d2

)

₌ 2 1 0 ε _{E (Ad)}2 ∴ Energy density (u) = energy per unit

Volume =

2 1

0 ε _E2

If dielectric is introduced then U =

2

1 _K

0 ε _E2

This energy is stored in a capacitor in the electric field between its plates.

FORCE ON A DIELECTRIC IN A CAPACITOR :

a b 1 2 3 4

IIT- PHY - CP

Let us consider a small displacement dx of the dielectric as shown in figure, keeping the net force on it always zero.

∴ Welectrostatic +

W

_F = 0

[Where

W

_F denotes the work done in displacement dx] ∴

W

F = - Welec. = ∆U ⇒ - F dx = 2 Q2 _d     C 1 = 2₂ C 2 Q − dc ⇒ F= 2 2 C 2 Q dx dc ₌ 2 1 ₂ V dx dc

[Considering capacitor has battery connected to it, i.e., V = Q/C ]

ILLUSTRATION : 04

Two capacitors of capacitances 20pf and 50pf are connected in series with a 6-volt battery, find

(A) The potential difference across each capacitor (B) The energy stored in each capacitor

SOLUTION :

(A) Equivalent capacitance C =

2 1 2 1

C

C

C

C

+

= 50 20 20 x 50 + = 7 100 pF ∴ Charge on C₁ = charge on C₂ = 7 100 x 6 = 7 600 pC

∴ Potential difference across

C

₁

(

=

50

pC

)

=

50 x 7 600 = 1.71 V and across

C

₂

=

(

20

pC

)

= 20 x 7 600 = 4.28 V (B) Energy in C₁ = E₁ = 2 1 x 50x

(

1.71

)

2 = 73.5 pJ Energy in C₂ = E₂ = 2 1 _{x 20 x}

₍

₎

₂ 28 . 4 = 184 pJ ILLUSTRATION : 05

A 5µF capacitor is charged to 12 volt. The positive plate of this capacitor is now connected to the negative terminal of a 12 V battery and vice versa. Calculate the heat developed in the connecting wires.

SOLUTION :

When capacitor is connected with battery the charge appears on one plate be Q = CV and - Q on the other plate. If the capacitor is now disconnected and connected to the same battery again with opposite polarity then - Q appear on first plate and + Q on second plate.

dx

x

F

20pF 50pF 6V

IIT- PHY - CP

∴ Total charge flown from battery is 2Q ∴W= charge x potential = 2QV ∴ Q = CV

∴ W = 2C_V2

⇒ W = 2x5x₁₀−6 _x

_{( )}

₁₂ 2

= 1.44 mJ

ILLUSTRATION : 06

A capacitor stores 50µC charge when connected across a battery. When the gap between the plates is filled with a dielectric, a charge of 100µC flows through the battery. Find the dielectric constant of the material inserted.

SOLUTION :

Initial charge = 50µC =Q1 Amount of charge flows = 100µC

∴ With dielectric total charge = (50+100) =150µC Initial capacity C = V Q ₌ V C 50µ Final capacity _C' ₌ V Q' ₌ V C 150µ K = C C' ₌ V / 50 V / 150 _{= 3} ILLUSTRATION : 07

In the above circuit, find the potential difference across AB.

SOLUTION :

Let us mark the capacitors as 1, 2, 3 and 4 for identification. As is clear, 3 and 4 are in series, and they are in parallel with 2. The 2,3, 4 combination is in series with 1.

,

f

4

C

C

C

.

C

C

4 3 4 3 34

=

₊

=

µ

C2,3,4 =8+4=12µf , f 8 . 4 12 8 12 8 C_eq = µ + × = q =CeqV=4.8×10=48µC The 'q' on 1 is 48µC, thus 6V c q V₁= = _      ₌ µ µ = 6V. F 8 c 48 V₁

⇒

VPQ =10−6=4V

By symmetry of 3 and 4, we say, VAB =2V.

ILLUSTRATION : 08

q

IIT- PHY - CP

What is V_A −V_B in the arrangement shown? What is the condition such that

0 V V_A− _B =

SOLUTION :

Let charge be as shown

(Capacitors in series have the same charge) Take loop containing C₁, C₂ and

E

0 E C q C q 2 1 + − = ⇒













+

=

2 1 2 1

C

C

C

C

E

q

From loop containing C3,C4and E Similarly, 0 E C ' q C ' q 4 3 + − = ⇒

_











+

=

4 3 4 3

C

C

C

C

E

'

q

Now, 4 2 B A C ' q C q V V − = − =

_











+

−

+

_{2 3} 3 ₄ 1 1

C

C

C

C

C

C

.

E

(

) (

)



_









+

+

−

=

−

4 3 2 1 2 3 4 1 B A

V

E

.

_C

C

.

C

_C

_.

_C

C

.

C

_C

V

For VA −VB =0 ⇒ C₁C₄ =C₂C₃ =0 or 4 3 2 1

C

C

C

C

=

ILLUSTRATION : 09

A 8µF capacitor C1 is charged to V = 120volt. The 0

charging battery is then removed and the capacitor is connected in parallel to an uncharged 4µF capacitor C2.

(A) What is the potential difference V across the combination? (B) What is the stored energy before and after the switch S is thrown ?

SOLUTION :

(A) Let q₀be the charge on C₁ initially Then q₀ =C₁V₀ when C₁ is connected to

2

C in parallel, the charge q0 is distributed between C1and C2. Let q1and q2 be the

charges on C₁and C₂ respectively. Now let V be the potential difference across each condenser.

S

1

C

C

2 0

V

IIT- PHY - CP Now q₀ =q₁+q₂ or C₁V₀ =C₁V+C₂V

(

120

V

)

F

4

F

8

F

8

V

C

C

C

V

₀ 2 1 1

µ

+

µ

µ

=

+

=

∴

= 80volt. (B) Initial energy stored

2 1 V C 2 1 U₀ = _{1 0}2 = (8 x 10-6_{) (120)}2 = 5 .76 x 10-2_joule

Final energy stored

U =

1

C V

₁ 2

1

C V

₂ 2

2

+

2

= 2 1 (8 x 10-6_{) (80)}2 ₊ 2 1 (4 x 10-6_{) (80)}2 = 3.84 x 10-2_joule.

Final energy is less than the initial energy. The loss of energy appears as heat in connecting wires.

ILLUSTRATION : 10

From the given figure find the value of the capacitance C if the equivalent capacitance between points A and B is to be 1µF. All the capacitances are in µF.

SOLUTION :

The capacitors C3 and C4 are in parallel, therefore

their resultant capacity C8 is 4. The capacitors C5 and C6

are in series, therefore, their resultant capacity C9 is 4.

These are shown in figure (A)

Now the capacitor C2 and C8 are in series. Their

resultant capacity C is 10

3 8

. Capacitors C₇and C are 9

in parallel. Their resultant capacity C11is 8. These are

shown in figure. (B)

C1 and C11 are in series. The equivalent capacitance is 8/9.

The parallel combination of 8/3 and 8/9 gives a resultant capacitance 32/9 as shown in figure. (C)

32 9 C 1 1 1_{= +} ∴ or 32 23 C 1 ₌ B A 2 C 8 C 1 C 7 C 9 C 1 8 4 4 4 C

( )

A

IIT- PHY - CP F 23 32 C= µ ∴ ILLUSTRATION : 11

Five identical conducting plates 1, 2, 3, 4 and 5 are fixed parallel to and equidistant from each other as shown in figure. Plates 2 and 5 are connected by a conductor while 1 and 3 are joined by another conductor. The junction of 1 and 3 and the plate 4 are connected to a source of constant e.m.f.

V

₀. Find

(i) The effective capacity of the system between the terminals of the source

(ii) The charge on plates 3 and 5

Given d = distance between any two successive plates and A = area of either face of each plate.

SOLUTION :

(i) The equivalent circuit is shown in figure (B). The system consists of four capacitors i.e., C12, C32 C34and C54. The capacity of each capacitor is 0 C0

d A K =      ε The

effective capacity across the source can be calculated as follow:

The capacitors C12 and C32 are in parallel and hence their capacity is C0 + C0 = 2C0.

The capacitor C54 is in series with effective capacitor of capacity 2C0. Hence the resultant

capacity will be 0 0 0 0

2C

C

C

2

x

C

+

Further

C

₃₄is again in parallel. Hence the effective capacity =

d

A

K

3

5

C

3

5

C

2

C

2C

x

C

C

_{0 0} 0 0 0 0 0

+

₊

=

=

ε

(ii) Charge on the plate 5 = charge on the upper half of parallel combination

d AV K 3 2 C 3 2 V Q_{5 0 0}= ε0 0      = ∴

Charge on plate 3 on the surface facing 4

d

AV

K

C

V

_{0 0}

=

ε

0 0

Charge on plate 3 on the surface facing 2

= [Potential difference across (3 - 2)] C0

=

d

AV

K

C

C

C

C

V

3

2

0 0 0 0 0 0 0

₊

=

ε

IIT- PHY - CP

d

AV

K

d

AV

K

Q

3

0 0 0 0 3

+

ε

ε

=

∴

= 0 0 ₀ V₀ d A K 3 4 3 1 1 d AV K ε =     + ε ILLUSTRATION : 12

(A) Find the effective capacitance between points X and Y in the given figure. Assume that C₂ =10µF and other capacitors are 4µF each.

(B) Find the capacitance of a system of identical capacitors between points A and B shown in figure.

SOLUTION :

(A) The circuit is redrawn in figure as the two arms are balanced, no current flows through C₂, C₃and C₄are in series, hence their equivalent capacitance = 2µF Similarly the equivalent capacitance of C₁and C₅= 2µF. Corresponding to points X and Y these two are in a parallel combination. Hence the effective capacitance between X and Y is 2 + 2 = 4µF.

(B) The arrangement of capacitors shown in figure is equivalent to the arrangement shown in figure. The arrangement is connected in parallel. Hence equivalent capacitance C is given by C = C1 + C2 + C3

WORKED OUT OBJECTIVE PROBLEMS :

EXAMPLE : 01

A parallel plate capacitor is connected across a 2V battery and charged. The battery is then disconnected and glass slab is introduced between the plates. Which of the following pairs of quantities decrease?

(A) Charge and potential difference (B) potential difference and energy stored

(C) energy stored and capacitance (D) capacitance and charge

Ans: (B) SOLUTION :

The introduction of a dielectric slab increases the capacitance. The charge remains unchanged. Potential difference and energy stored decreases.

EXAMPLE : 02

Three capacitors of capacitances 3µF, 9µF and 18µF are connected once in series and then in parallel. The ratio of equivalent capacitances in the two cases(CS/CP) will be

(A) 1 : 15 (B) 15 : 1 (C) 1 : 1 (D) 1 : 3

IIT- PHY - CP SOLUTION : F 30 18 9 3 C_P = + + = µ 2 1 18 1 9 1 3 1 C 1 S = + + =

∴

C_S =2µF Now

15

1

F

30

F

2

C

C

P S

₌

µ

µ

=

EXAMPLE : 03

A number of capacitors each of capacitance 1µF and each one of which get punctured if a potential difference just exceeding 500volt is applied, are provided. Then an arrangement suitable for giving a capacitor of capacitance 2µF across which 3000 volt may be applied requires at least

(A) 18 component capacitors (B) 36 component capacitors (C) 72 component capacitors (D) 144 component capacitors

Ans: (C) SOLUTION :

Number of capacitors required in series = 6 500

3000 ₌

The capacitance of series combination. = F 6

1_µ

To obtain a capacitor of 2µF, we should use 12 such combinations. ∴Total number of capacitors required = 12 x 6 = 72

EXAMPLE : 04

A capacitor of capacitance 1µF withstands a maximum voltage of 6kV, while another capacitor of capacitance 2µF, the maximum voltage 4kV. If they are connected in series, the combination can withstand a maximum of

(A) 6kV (B) 4kV (C) 10kV (D) 9kV

Ans: (D) SOLUTION :

When the two condensers are connected in series.

3 2 1 2 1 x 2 C = + = µF and E 3 2 Q=

The potential of condenser C₁is given by

kV 6 E 3 2 C Q V 1 1= = < ⇒E<9kv E 12KV 3 E c Q v 2 2 = = < <

IIT- PHY - CP EXAMPLE : 05

Seven capacitors each of capacitance 2µF are to be connected to obtain a capacitance of       11 10

µF. Which of the following combination is possible. (A) 5 in parallel 2 in series (B) 4 in parallel 3 in series (C) 3 in parallel 4 in series (D) 2 in parallel 5 in series

Ans: (A) SOLUTION :

5 capacitors in parallel gives 5 x 2 µF = 10µF capacity. Further, two capacitors in series gives a capacity 1µF. When the two combinations are connected in series, they give a resultant capacitance       +1 10 1 x 10 =       11 10 µF. EXAMPLE : 06

Condenser A has a capacity of 15µF when it is filled with a medium of dielectric constant 15. Another condenser B has a capacity 1µF with air between the plates. Both are charged separately by a battery of 100V. After charging, both are connected in parallel without battery and the dielectric material being removed. The common potential now is

(A) 400V (B) 800V (C) 1200V (D) 1600V

Ans: (B) SOLUTION :

Charge on capacitor A is given by q₁ =C₁ x V

= (15 x 10-6_{) (100) = 15 x 10}-4_C

Charge on capacitor B is given by q2 = C2 x V

= (1 x 10-6_{) (100) = 10}-4_C

Capacity of condensers A after removing dielectric

1 15 10 x 15 K C ' C -6 1 ₌         =       = µF

Now when both condenser are connected in parallel their capacity will be 1µF + 1µF = 2µF Common potential V = C q =

(

)

800V 10 x 2 ) 10 x 1 ( 10 x 15 6 -4 -4 -= + _. EXAMPLE : 07

Two capacitors 2µF and 4µF are connected in parallel. A third capacitor of 6 µF capacity is connected in series. The combination is then connected across a 12V battery. The voltage across 2µF capacity is

IIT- PHY - CP (A)2V (B) 6V (C) 8V (D) 1V Ans: (B) SOLUTION :

Resultant capacitance of condensers of capacity 2µF and 4µF when connected in parallel. 6 4 2 ' C= + = µF

This is connected in series with a capacitor of capacity 6µF in series. The resultant capacity C is given by 3 1 6 1 6 1 C 1 = + = or C=3µF Charge on combination q = (3 x 10-6_{) x (12) = 36 x 10}-6_C

Let the charge on 2µF capacitor be q₁, then

4

q

q

2

q

₁

₌

−

_{1 or} 3 q q₁ = ∴q₁ =12 x 10-6_C

Now potential across 2µF condenser = = _-6 = -6 6 -1 10 x 2 10 x 12 10 x 2 q 6V EXAMPLE : 08

The capacitance of the system of parallel plate capacitor shown in the figure is (A)

₍

₎

d

A

A

A

A

2

2 1 2 1 0

+

ε

(B)

₍

₎

d

A

A

A

A

2

1 2 2 1 0

−

ε

(C)

d

A

₁ 0

ε

(D)

d

A

₂ 0

ε

Ans : (C) SOLUTION :

Since the electric field between the parallel charge plates is uniform and independent of the distance, neglecting the fringe effect, the effective area of the plate of area A2 is A1. Thus the

capacitance between the plates is

d

A

C

=

ε

0 1 ∴ (C)

EXAMPLE : 9

The charge flowing across the circuit on closing the key K is equal to (A) CV (B) CV 2 (C)2CV (D) Zero SOLUTION : 1

A

2

A

d

1

A

2

A

d

ε

0

σ

+

σ

−

E

IIT- PHY - CP

When the key K is kept open the charge drawn from the source is Q=C'V

Where C' is the equivalent capacitance given by

2 C ' C = Therefore Q = V 2 C      

Whey the key K is closed, the capacitor 2 gets short circuited and the charge in the circuit

Q

₁

=

CV

∴ Charge flowing is

Q Q

₁

Q

C

V

2

∆ =

− =

∴ (B) EXAMPLE : 10

The figure shows a spherical capacitor with inner sphere earthed. The capacitance of the system is

(A)

a

b

ab

4

₀

−

πε

(B) a b b 4 ₀ 2 − πε (C)4πε0

(

b+a

)

(D) None of these SOLUTION :

Let V be potential of the outer sphere. Thus we can consider two capacitors, one between the outer sphere and inner sphere and the other between outer sphere and infinity.

Thus, a b ab 4 C_{1 0} − πε = b 4 C₂ = πε₀ b 4 a b ab 4 C ₀ + πε₀ − πε = ⇒ a b b 4 C 2 0 − πε = ∴ (B) EXAMPLE : 11

A capacitor of capacitance C is charged to a potential difference V from a cell and then disconnected from it. A charge +Q is now given to its positive plate. The potential difference across the capacitor is now

(A) V (B) C Q V+ (C) C 2 Q V+ (D) , C Q V− if V < CV SOLUTION : a b a b • 1 C C2

IIT- PHY - CP

In the figure given below, left X and Y be positive and negative plates. After charging from the cell, the inner faces of X and Y have charges ±CV, as shown in (A). The outer surfaces have no charge.

When charge Q is given to X, let the inner faces of X and Y have charges ±q Then, by the principle of charge conservation, the outer faces have charges

(

Q+CV−q

)

for X and

(

q−CV

)

for Y, as shown in (B). Now, the outer faces must have equal charges. ∴ Q+CV−q=q−CV or 2q=2CV+Q or 2 Q CV q= + Potential difference C 2 Q V C q _{= +} ∴ (C) EXAMPLE : 12

In an isolated parallel - plate capacitor of capacitance C, the four surfaces have charges Q1,Q2,Q3 and Q4 as

shown. The potential difference between the plates is (A)

C

2

Q

Q

Q

Q

₁

+

₂

+

₃

+

_{4 (B)}

C

2

Q

Q

₂

+

₃ (C)

C

2

Q

Q

₂

−

_{3 (D)}

C

2

Q

Q

₁

+

₄ SOLUTION :

Plane conducting surfaces facing each other must have equal and opposite charge densities. Here, as the plate areas are equal, Q2 = - Q3

The charge on a capacitor means the charge on the inner surface of the positive plate-(in this case Q2)

Potential difference between the plates = charge on the capacitor ÷ capacitance. ∴ Potential difference =

(

)

C

2

Q

Q

C

2

Q

Q

C

2

Q

2

C

Q

_{2 2 2 2 2}

−

₃

=

−

−

=

=

∴ (C) * * * 1 Q Q3 2 Q Q4

IIT- PHY - CP

SINGLE ANSWER OBJECTIVE TYPE QUESTIONS :

LEVEL - 1 :

1. A capacitor of capacitance C is charged to potential difference V0 from a cell and

then disconnected from it. A charge +Q is now given to its positive plate. The potential difference across the capacitor is now

A) V0 B) V0 + (Q/C) C) V0 + (Q/2C) D) V0 - (Q/C)

2. A dielectric of dielectric constant 3 fills up three fourths of the space between the plates of a parallel plate capacitor. The percentage of energy stored in the dielectric is

A) 25% B) 50% C) 75%

D) 100%

3. A parallel plate capacitor having capacitance C0 is connected to a battery of emf E.

It is then disconnected from the battery and a dielectric slab of dielectric constant k completely filling the air gap of the capacitor is inserted in it. If ∆U indicates the change in energy, then

A) ∆U = 0 B) ∆U = 2 1 ∈0E2 (k - 1) C) ∆U = 2 1 ∈0E2       − k 1 1 D)∆U 2 1 ∈0E2       −1 k 1

4. The work done in increasing the voltage across the plates of the capacitor from 5V to 10V is W. The work done in increasing the voltage from 10V to 15V will be

A) W B) 4/3 W C) 5/3 W D) 2W

5. A parallel plate capacitor of plate area A and plate separation d is charged to potential difference V and then the battery is disconnected. A slab of dielectric constant K is then inserted between the plates of the capacitor so as to fill the space between the plates. If Q, E and W denote respectively, the magnitude of charge on each plate, the electric field between the plates (after the slab is inserted), and work done on the system, in questions, in the process of inserting the slab, then A) _Q=∈0 AV d B) ∈ = 0 KAV Q d C) E = Kd V D) ∈   = _ − _÷   2 0 AV 1 W 1 2d K

6. When the capacitance in an oscillator circuit of frequency f is increased nine times, the frequency of the oscillator is reduced to:

A) f/9 B) f/6 C) f/4 D) f/3

7. 64 small drops of water having the same charge & same radius are combined to form one big drop. The ratio of capacitance of big drop to small drop is:

A) 4 : 1 B) 1 : 4 C) 2 : 1

IIT- PHY - CP

8. A parallel plate condenser is connected to a battery. The plates are pulled apart with a uniform speed. If x is the separation between the plates, then the time rate of charge of the electrostatic energy of the condenser is proportional to

A) x2 _{B) x C) 1/x D) 1/x}2

9. A spherical condenser has inner and outer spheres of radii a and b respectively. The space between the two is filled with air. The difference between the capacities of two condensers formed when outer sphere is earthed and when inner sphere is earthed will be

A) zero B) 4π∈0 a C) 4π∈0 b D) 4π ∈0 a [b/(b -

a)]

10. The sphere shown in the figure are connected by a conductor. The capacitance of the system is:

A) 4 πε0 a b ab − B) 4 πε0 a C) 4 π ε0 b D) 4πε0 a b a2 −

11. A dielectric slab of thickness d is inserted in a parallel plate capacitor whose negative plate is at X = 0 and positive plate is at X = 3d. The slab is equidistant from the plates. The capacitor is given some charge. As X goes from 0 to 3d.

A) the electric potential increases at first, then decreases and again increases B) the electric potential increases continuously

C) the direction of the electric field remains the same D) the magnitude of the electric field remains the same

12. Two similar conducting balls are placed near each other in air. The radius of each ball is r and the separation between the centers is d (d >> r). The capacitance of two balls system when they are connected by a wire is:

A) 8πε0r B) 4πε0r C) 4πε0r loge (r/d) D) 4π

loge ε0 (r/d)

13. A hollow sphere of radius 2R is charged to V volt and another smaller sphere of radius R is charged to V/2 volt. Then the smaller sphere is placed inside the bigger sphere without changing the net charge on each sphere. The potential difference between the two spheres would be

A) 3V/2 B) V/4 C) V/2 D) V

14. A capacitor is charged by using a battery which is then disconnected. A dielectric slab is then slipped between the plates which results in:

A) reduction of charges on the plates and increase of potential difference across the plates

B) increase in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates

C) decrease in the potential difference across the plates, reduction in stored energy, but no change in the charge on the plates

IIT- PHY - CP

D) none of the above

15. Force acting on a charged particle kept between the plates of a charged condenser is F. If one of the plates of the condenser is removed, force acting on the same particle will become:

A) zero B) F/2 C) F D) 2F

16. A parallel plate air capacitor is connected to a battery. The quantities charge, voltage, electric field and energy associated with this capacitor are given by Q0, V0,

E0 and U0 respectively. A dielectric slab is now introduced to fill the space between

the plates with battery still in connection. The corresponding quantities now given by Q, V, E and U are related in the previous ones as:

A) Q > Q0 B) V > V0 C) E > E0 D) U > U0

17. Figure shows two capacitors connected in series and7joined to a battery. The graph shows the variation in potential as one moves from left to right on the branch containing the capacitors:

A) C1 > C2 B) C1 = C2

C) C1 < C2

D) the information is not sufficient to decide the relation between C1 and C2

LEVEL - II :

1. Identical charged 103 _{oil drops each of radius 0.9nm are combined to form a single}

drop. The electrical capacity of the drop is

1) 0.1 PF 2) 1 PF 3) 10 PF 4) 100 PF

2. A parallel plate condenser is charged end isolated. The distance between the plate is increased by 2mm and a dielectric slab of thickness 3 mm is introduced between the plates. If the potential difference between the plates remains same, the dielectric constant of the dielectric slab is

1) 2 2) 3 3) 4 4) 5

3. Two identical parallel plate condensers are connected in series. A cell of e.m.f of 20V is connected between their ends. A dielectric slab of constant 4 is placed between the plates of one of the condensers. The potential difference across condenser with dielectric slab is

1) 4V 2) 10V 3) 16V 4) 18V

4. Four identical parallel metal plates each of area A are placed with separation ‘d’ between adjacent plates as shown. The capacitance between the plates P and Q is 1) 3 0 A d ∈ 2) 2 0 A 3d ∈ 3) 2 0 A d ∈ 4) 0 A 2d ∈

5. Two spherical conductors of radii 3cm and 6cm are in contact. A charge 10-9 _is

given to them. The potential of the smaller sphere is

1) 67V 2) 33V 3) 50V 4) 100V

6. A parallel plate condenser is charged and isolated. The energy stored by the condenser is E. The separation of the plate is doubled and then the space is completely filled with a dielectric of constant 5. The energy stored by the

IIT- PHY - CP

condenser now is

1) E 2) 0.8E 3) 0.4E 4) 2.5E

7. Two condensers charged to potentials 50V and 80V are connected in parallel, the common potential is 60V. The capacities of the condensers are in the ratio of

1) 2 : 1 2) 1 : 2 3) 3 : 4 4) 4 : 3

8. The capacitor of 4µF charged to 50V is connected to another capacitor of 2µF charged to 100V. The total energy of the combination is

1) 4x10 J2 3 − ₂₎ 3_{x10 J}2 2 − _{3) 3 x10 J}−2 ₄₎ 8_{x10 J}2 3 −

9. The radii of two charged metal spheres are 5cm and 10cm both having same charge 75µc. If they are connected by a wire, the quantity of charge transferred through the wire is

1) 75µc 2) 50µc 3) 25µc

4) 15µc

10. Two identical capacitors have equivalent capacity of 2µF when they are connected in series. If they are connected in parallel and charged to a potential of 200V, the energy stored in the system is

1) 18 x 10-4 _{J 2) 18 x 10}-4 _{J 3) 0.16 J 4) 0.36}

11. The capacity of a parallel plate condenser is C. When half the space between the plates is filled with a slab of dielectric constant K as shown in the figure. If the slab is removed from the condenser, then the capacity of the condenser becomes

1) 2KC K 1+ 2)

(

)

K 1 C 2K + 3)

(

K 1 C

)

2K + 4) KC 2K 1+

12. Two parallel plate capacitors C and 2C are connected in parallel and charged to P.D ‘V’. The battery is then disconnected and the region between the plates of the capacitor ‘C’ is completely filled with a material of dielectric constant ‘3’. The P.D. across the capacitors now becomes

1) V 2) 3V 3) 3V/5 4) 4V/5

13. A condenser of capacity 10µF is connected between the terminals of a battery of potential difference 100V and is charged. The amount of work done by the battery to charge the condenser is

1) 0.05 J 2) 0.025 J 3) 0.1 J 4) 0.0125 J

14. Two condensers of capacities C1 = 5 µF – 100V and C2 = 10µF – 100V are connected

in series. The maximum potential difference that may be applied between them without damaging the condensers is

1) 200 V 2) 175 V 3) 150 V

4) 300 V

15. ‘n’ identical charged spheres combine together to form a large sphere. The ratio of the potential of small sphere to potential of large sphere is 1 : 9. The ratio of the energy stored in small sphere to the energy stored in the large sphere is

1) 1 : 3 2) 1 : 27 3) 1 : 81 4) 1 :

IIT- PHY - CP

16. The capacity of parallel plate condenser with air between the plates is 10µF. If the space between the plates is completely filled with two dielectric slabs each of thickness is equal to half of the separation between the plates, whose dielectric constants are 3 and 2. What will be the percentage change in capacity of the condenser?

1) 133.3 % 2) 140 % 3) 166.6 % 4) 280 %

17. The work done in charging a capacitor from 5V to 10V is W. The work done in

charging the

capacitor from 10V to 15V is

1) 4 W/3 2) 5 W/3 3) 2 W/5 4) 5 W/2

18. The voltages across C1 and C2 are in the ratio 2 : 3. When C2 is

completely filled with paraffin, the voltage ratio became 3 : 2. The dielectric constant of paraffin is

1) 2.25 2) 13/6

3) 27/8 4) 6

19. A parallel plate capacitor of capacity 5µF and plate separation

6cm is connected to a 1V battery and is charged. A dielectric of dielectric constant 4 and thickness 4cm is introduced into the capacitor. The additional charge that flows into the capacitor from the battery is

1) 2µC 2) 3µC 3) 5µC 4) 10µC

20. A metal sphere ‘A’ of radius a is charged to a potential ‘V’. What will be its potential if it is enclosed by a spherical conducting shell ‘B’ of radius b and the two are connected by a conducting wire?

1) V 2) (a/b) V 3) (b/a) V 4) zero

21. The capacity of a parallel plate capacitor is 5µF. When a glass plate of same area as the plates but thickness half of the distance between the plates is placed between the plates of the capacitor, its potential difference reduces to 2/5 of the original value. The dielectric constant of the glass is

1) 1.5 2) 2.5 3) 5 4) 2s Hint: 0 0 0 0 Qd V E d A Qd t 1 V 1 1 A d k = = ∈    = _ − _ − _÷ ∈    ⇒ 0 2 V t 1 1 1 3 1 1 ; 1 5 V d k 2 k 5     = = − _ − _{÷ } − _÷=    

22. Two spheres A and B of radii 4cm and 6cm are given charges of 80µC and 40µC. If they are connected by a wire, the amount of charge flowing from one to other is 1) 20µC from A to B 2) 16µC from A to B 3) 24µC from B to A 4) 32µC from A to B Hint: ∆Q = (C1 – C2)V =

(

) (

) (

) (

)

2 1 1 1 2 2 2 1 1 2 1 2 1 2 C C C V C V C C Q Q C C C C − + − + = + + i.e. ∆Q =

(

2 1

) (

1 2

)

1 2 r r Q Q 2 x 120 r r 10 − + = = + 24µC from B to A

23. Three capacitor of capacitance 10µF, 15µF, 20µF are in series with a cell. The charge drawn from the cell is 60µC. If they are connected in parallel with the same

IIT- PHY - CP

cell, then the charge drawn from the cell is

1) 385 µC 2) 485 µC 3) 585 µC 4) 685 µC Hint: p p p _p s s s q C E C ₄₅ q 6 x x13 585 C q = C E = C ⇒ = 60 = µ

24. The plates of a parallel plate capacitor are horizontal and parallel. A thin conducting sheet P is initially placed parallel to both the plates and nearer to the lower plate. From t = 0 onwards, the sheet P is moved at constant speed vertically upwards so that it is always parallel to the capacitor plates. At t = 20 milli seconds, it is nearer to the upper plates. Then during the time interval from t = 0 to t = 20 milli seconds, the capacity of the capacitor will

1) increase gradually 2) decrease gradually

3) remains constant 4) first increases and then decreases

Hint: Since the thickness of conducting plate is constant through out, capacity

remains constant

25. A capacitor is charged to 200V. A dielectric slab of thickness 4mm is inserted. The distance between the plates is increased by 3.2mm to maintain the same potential difference. Find the dielectric constant of the slab

1) 3 2) 4 3) 5 4) 6 Hint: V1 = V2; 1 2 2 1 0 0 Qd Q 1 1 d d d t 1 1 A A k k t −    = _ − _ − _÷⇒ = − ∈ ∈   

26. A capacitor is charged with a dielectric to V volts. If the dielectric of constant K is removed then ___ is true

a) capacity decreases by k times

b) electric field intensity decreased by k times c) potential increases by k times

d) charge increases by k times

1) a, b, c 2) a, b, c, d 3) a, c 4) b, d

27. The capacitance of a capacitor becomes 7/6 times its original value if a dielectric slab of thickness t = 2/3d is introduced in between the plates ‘d’ is the distance between the plates. The dielectric constant of dielectric field is

1) 14/11 2) 11/14 3) 7/11 4) 11/7 Hint: 0 0 C t 1 C 6 C 1 1 t 1 d k C 7 1 1 d k   = ⇒ − _ − _÷= =     − _ − _÷  

28. Between the plates of a parallel plate capacitor of capacity C, two parallel plates of the same material and area same as the plate of the original capacitor are placed. If the thickness of these plates is equal to I/5th _{of the distance between the plates of}

the original capacitor, then the capacity of the new capacitor is

1) 5/3 C 2) 3/5 C 3) 3C/10 4) 10C/3 Hint: d1 = d, d2 = 3 5d, t = 2 5d & 2 1 2 1 2 C d 5 5 C C C =d = ⇒3 =3

29. A parallel plate capacitor of capacity Co is charge to a potential V0.

A) The energy stored in the capacitor when the battery is disconnected and the plate separation is double is E1

IIT- PHY - CP

B) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is E2, the E1/E2 value is

1) 4 2) 3/2 3) 2 4) ½

30. A capacitor of capacity of 10µF is charged to 40V and a second capacitor of capacity 15µF is charged to 30V. If they are connected in a parallel the amount of charge that flows from the smaller capacitor to higher capacitor in µC is

1) 30 2) 60 3) 200 4) 250 Hint: V = 1 1 2 2 1 2 C V C V & C C + + ∆Q = (V1 – V)C1 = 60µC

31. A parallel plate capacitor of capacity 100µF is charged by a battery of 50 volts. The battery remains connected and if the plates of the capacitor are separated so that the distance between them becomes half the original distance, the additional energy given by the battery to the capacitor in joules is

1) 125 x 10-3 _{2) 12.5 x 10}-3 _{3) 1.25 x 10}-3 _{4) 0.125 x 10}-3

Hint: ∆V = ½(C2 – C1)V2 & C2 = 2C

32. A parallel plate capacitor of capacity 5µF and plate separation 6cm is connected to a I volt battery and is charged. A dielectric of dielectric constant 4 and thickness 4cm introduced into the capacitor. The additional charge that flows into the capacitor from the battery is

1) 2µC 2) 3µC 3) 5µC 4) 10µC Hint:

Q

₁

=

C

₀V = 5µC; Q2 = C2V = 0 C V t 1 1 1 d k   − _ − _÷   = 10µC; ∆Q = Q2 – Q1

33. A 20F capacitor is charged to 5V end isolated. It is then connected in parallel with an uncharged 30F capacitor. The decrease in the energy of the system will be

1) 25 J 2) 100 J 3) 125 J 4) 150 J Hint: 1 1 1 1 12 2 1 1 2 1 2 C V 1 C V & U C V U U C C 2 C C = = ⇒ ∆ = + +

34. A dielectric of thickness 5cm and dielectric constant 10 is introduced in between the plates of a parallel plate capacitor having plate area 500 sq cm and separation between plates 10cm. The capacitance of the capacitor is (∈0=8.8 x 10-12 SI units)

1) 8 PF 2) 6PF 3) 4PF 4) 20PF Hint: C = 0 A 1 d t 1 k ∈   − _ − _÷  

35. A 4µF capacitor is charged by 200V battery. It is then disconnected from the supply and is connected to another uncharged capacitor of 2µF capacity. The loss of energy during this process is _____

1) 0 2) 5.33 x 10-2 _{3) 4 x 10}-2 _{4) 2.67 x 10}-2 Hint: ∆U = 1 2

(

1 2

)

2 2 1 1 1 2 1 2 C C C V 1 V V ; V 2 C +C − =C +C

IIT- PHY - CP

36. Energy E is stored in a parallel plate capacitor C1. An identical uncharged capacitor

C2 is connected to it kept in contact with it fro a while and then disconnected. The

energy stored in C2 is ___

1) E/2 2) E/3 3) E/4 4) 0

Hint: U1 = 1C V₁ 2

2 , Potential on second capacitor in contact with the first one V2 =

V/2; U2 =

2

1 V E

C

2   = ÷_{ }2 4

37. If the capacity of a spherical conductor is 1PF, then its diameter is

1) 9 x 10-15_{m 2) 9 x 10}-3_{m 3) 9 x 10}-5_{m 4) 18 x 10}-7_m

38. A 700PF capacitor is charged by a 50V battery. The electrostatic energy stored by it is

1) 17 x 10-5 _{J 2) 13.6 x 10}-9 _{J 3) 9.5 x 10}-9 _{J 4) 8.75 x 10} -7 _J

39. Two equal capacitors are first connected in parallel and then in series. The ratio of the total capacities in the two cases will be

1) 2 : 1 2) 1 : 2 3) 4 : 1 4) 1 : 4

40. Two condensers of capacity 2C and C are joined in parallel and charged up to potential V. The battery is removed and the condensor of capacity ‘C’ is filled completely with a medium of dielectric constant K. The potential difference across the capacitors will now be

1) 3V K 2) 3V K 2+ 3) V K 2+ 4) V K

41. Two condensers C1 and C2 in a circuit are joined

as shown in figure. The potential at A is V1 and

that of B is V2. The potential of point D will be

1) 1 1 2 2 1 2 C V C V C C + + 2) 1 2 2 1 1 2 C V C V C C + + 3) 1 2 V V 2 + 4) 1 2 2 1 1 2 C V C V C C − +

42. A number of capacitors are connected as shown in figure. The equivalent capacity is given by

1) nC 2) n(n + 1)C

3) 2n(n + 1)C 4) n n 1 C

(

)

2

+

43. Three capacitors of capacitances 3µF, 10µF and 15µF are connected in series to a voltage source of 100V. The charge on 15µF is

1) 50µC 2) 100µC 3) 200µC 4) 280µC

44. In a parallel plate capacitor of capacitance ‘C’, a metal sheet is inserted between the plates parallel to them. If the thickness of the sheet is half of the separation between the plates, the capacitance will be

IIT- PHY - CP

1) C/2 2) 3C/4 3) 4C 4) 2C

45. A 10 micro farad capacitor is charged to 500 V and then its plates are joined together through a resistance of 10 ohm. The heat produced in the resistance is

1) 500 J 2) 250 J 3) 125 J

4) 1.25 J

46. A capacitor is charged to 200 volt. It has a charge of 0.1C. When it is discharged, energy liberated will be

1) 1 J 2) 10 J 3) 14 J 4) 20 J

47. Half of the separation between two parallel plates of a capacitor is filled with a dielectric medium. The capacitance of the capacitor becomes 5/3 times its original value with full space dielectric. The dielectric constant of the medium K is

1) K = 2 2) K = 3 3) K = 4

4) K = 5

48. A 10µF capacitor is charged to a potential difference of 50 V and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20 V. The capacitance of second capacitor is

1) 10 µF 2) 20 µF 3) 30 µF

4) 15 µF

49. Capacity of a spherical capacitor having two spheres of radii a and b (a > b) separated by a medium of dielectric constant K is given by

1) Kab a b− in SI system 2) 0 4 Kab a b π∈ − in CGS system 3) 4

₍

0 Kab

₎

a b π∈ − in SI system 4)

(

)

K a b ab − in CGS system

50. A capacitor of capacity ‘C’ has charge Q. The stored energy is W. If the charge is increased to 2Q, the stored energy will be

1) 2W 2) W/2 3) 4W 4) W/4

51. A D.C. potential of 100 volt is connected to the combination as shown in figure. The equivalent capacity between A and B will be equal to

1) 40 µF 2) 20 µF

3) 30 µF 4) 10 µF

52. The capacitance of four plates, each of area A arranged as shown in figure as 1) 2 0 A d ∈ 2) 3 0 A d ∈ 3) 4 0 A d ∈ 4) 5 0 A d ∈

53. Identify the wrong statement of the following

a) the resultant capacity ‘C’ is less than the capacitance of smallest capacitor in series combination

b) the resultant capacity ‘C’ is greater than greatest capacitance in parallel combination

IIT- PHY - CP

c) In series combination, charge on capacitor plates is inversely proportional to capacitance of capacitor

1) a is wrong 2) c is wrong 3) b is wrong 4) all are wrong 54. A parallel plate capacitor of area A, plate separation d with the

electric capacity C0 is filled with three different dielectric

materials with constants K1, K2 and K3 as in the figure. If these

three are replaced by a single dielectric, its dielectric constant K is ___ 1) 1 2 3 1 1 1 1 K= K +K +2K 2) _{1 2 3} 1 1 1 K= K +K +2K 3) 1 2 3 1 2 K K 1 2K K K K + = + + 4) 1 3 2 3 1 3 2 3 K K K K K K K K K = + + +

55. Four metallic plates, each with a surface area of one side A, are placed at a distance d apart. The outer plates are connected to terminal X and the inner plates to terminal Y. The capacitance of system between X and Y is

1) ε0A/d 2) 2ε0A/d 3) 3ε0A/εd 4) 4ε0A/d

56. Four metallic plates, each with a surface area of one side A are placed at a distance

d apart, these plats are connected as shown in figure.

The capacitance of the system between X and Y is 1) ε0A/d 2) 2ε0A/d

3) 3ε0A/d 4) 4ε0A/d

57. An infinite ladder is made as shown in figure using capacitors C1 = 1 µF and C2 = 2µF. The equivalent

capacitance of the ladder, in µF is

1) 1 2) 2

3) 0.75 4) 0.5

58. In the circuit segment shown VA - VB = 19V. The p.d. across 3 µF capacitor is

1) 7V 2) 8V

3) 23V 4) 4V

59. A parallel plate capacitor with a dielectric constant K = 3 filling the space between the plates is charged to a potential difference V. The battery is then disconnected and the dielectric slab is withdrawn and replaced by another dielectric slab having K = 2. The ratio of energy stored in the capacitor before and after replacing the dielectric slab by new one is

1) 3 : 2 2) 9 : 4 3) 4 : 9 4) 2 : 3

60. Two parallel capacitors of capacitance C and 2C are connected in parallel and charged to a potential difference V. The battery is then disconnected and the capacitor C is completely filled with a material of dielectric constant k. The potential difference across the capacitors is now

1) 2V/k 2) 3V/k 3) 3V/(k+2) 4) 2V/(k+3)

61. Initially the capacitors C1 and C2 shown in figure have equal

capacitances. If a dielectric plate (k = 2) is introduced in capacitor C2, then potential difference across its plates and

IIT- PHY - CP

charge

1) both will decrease 2) both will increase 3) p.d. will increase but charge will decrease 4) p.d. will decrease but charge will increase

62. Five identical capacitor plates, each of area A, are arranged such that adjacent plates are at a distance d apart. The plates are connected to a source of emf V as shown in figure. The charge on plate 1 is q and that on 4 is q', where

1) q' = q 2) q' = 2q

3) q' = -2q 4) q' = 3q

63. For the circuit shown in figure which of the following statements is true

1) With S1 closed, V1 = 15V, V2 = 20V

2) With S3 closed, V1 = V2 = 25V

3) With S1 and S2 closed, V1 = V2 = 0

4) With S1 and S3 closed, V1 = 30V, V2 = 20V

64. Two identical capacitors, having the same capacitance C. One of them is charged to a potential V1 and the other to V2. The negative ends of the capacitor are

connected together. When the positive ends are also connected, the decrease in energy of the combined system is

1) C(V V ) 4 1 2 2 2 1 − 2) C(V V ) 4 1 2 2 3 1 + 3) C(V1 V2)2 4 1 _{− 4)} 2 2 1 V ) V ( C 4 1 ₊

65. Each edge of a cube (figure) made of wire contains a capacitor of capacitance C. Find the effective capacitance of this bank of capacitors between opposite corners A and G.

1) 5C/6 2) 4C/3

3) 3C/4 4) 6C/5

66. Consider the situation shown in figure. The capacitor A has charge q on it whereas B is uncharged. The charge appearing on the capacitor B a long time after switch is closed is

1) zero 2) q/2 3) q 4) 2q

67. A parallel plate capacitor of capacitance C is connected to a battery and is charged to a potential difference V. Another capacitor of capacitance 2C is similarly charged to potential differences 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the negative terminal of one is connected to the negative terminal of the other. The final energy of the configuration is

1) zero 2) (3/2)

CV

2 3) (25/6)

CV

2 4) (9/2)

CV

2

68. You are given thirty two capacitors each having capacity 4 µF. How do you connect all of them to prepare a composite capacitor having capacitance 8 µF?

1) 4 Condensers in series 8 such groups in parallel 2) 2 Condensers in series and 16 such groups in parallel

3) 8 Condensers in series and 4 such groups in parallel 4) All of them in series. 69. Find out the effective capacitance between points P and Q. It will be

IIT- PHY - CP

1) 9 µF 2) 4.5 µF 3) 1 µF 4) 6 µF

70. An uncharged parallel plate capacitor having a dielectric of constant K is connected to a similar air filled capacitor charged to a potential V. The two share the charge and the common potential is V'. The dielectric constant K is 1) V V V V + ′ − ′ 2) V V V ′ − ′ 3) V V V′− 4) V V V ′ ′ −

71. Two condensers each having capacitance C and breakdown voltage V are joined in series. The capacitance and the breakdown voltage of the combination will be

1) 2C and 2V 2) C/2 and V/2 3) 2C and V/2 4) C/2 and 2V

72. Two identical metal plates are given positive charges Q1 and Q2 (< Q1) respectively. If they are now

brought close together to form a parallel plate capacitor with capacitance C, the potential difference between them is 1) C 2 ) Q Q ( 1+ 2 ₂₎ C ) Q Q ( 1+ 2 ₃₎ C ) Q Q ( 1− 2 ₄₎ C 2 ) Q Q ( ₁− ₂

73. Three very large plates are given charges as shown in the figure. If the cross sectional area of each plate is the same, the final charge distribution on plate C is:

1) +5 Q on the inner surface, +5 Q on the outer surface 2) +6 Q on the inner surface, +4 Q on the outer surface 3) +7 Q on the inner surface, +3 Q on the outer surface 4) +8 Q on the inner surface, +2 Q on the outer surface

74. Two identical sheets of metallic foil are separated by d and capacitance of the system is C and is charged to a potential difference E. Keeping the charge constant, the separation is increased by l. Then the new capacitance and potential difference will be:

1)

d

A

0

ε

,E 2) ) d ( A 0 l + ε ,E 3)       + + ε d 1 , ) d ( A 0 l l E 4) _     + ε d 1 , d A 0 l _E

75. n conducting plates are placed face to face as shown in

figure. Distance between any two plates is d. Area of the plates is A,

A A A

, , ...

_{(n 1)}

A

2 4 8

2

− . The equivalent

capacitance of the system is 1)

d

2

A

n 0

ε

2) d ) 1 2 ( A n0₋ ε 3) d ) 2 2 ( A n0₋ ε 4) d ) 1 2 ( A n 0 − ε

76. The adjoining figure shows two identical parallel plate capacitors connected to a battery with switch S close(4). The switch is now opened and the plates are filled with a dielectric of dielectric constant 3. The ratio of the total electrostatic energy stored in

both the capacitors before and after the introduction of the dielectric is

IIT- PHY - CP

1) 2 : 5 2) 3 : 5 3) 5 : 2 4) 5 : 3

77. A parallel plate capacitor has two layers of dielectric as shown in figure.

This capacitor is connected across a battery, then the ratio of potential

difference across the dielectric layers

1) 4/3 2) 1/2

3) 1/3 4) 3/2

78. Five conducting plates are placed parallel to each other. Separation between them is d and area of each plate is A. Plate number 1, 2 and 3 are connected with each other and at the same time through a cell of emf E. The charge on plate number 1 is

1) Eε0A/d 2) Eε0A/2d 3) 2 Eε0A/d 4) zero

79. A capacitor of capacity C1, is charged by connecting it across a battery of emf V0. The battery is then

removed and the capacitor is connected in parallel with an uncharged capacitor of capacity C2. The

potential difference across this combination is:

1) 0 2 1 2

_V

C

C

C

+

2)

C

₁ 1

C

₂

V

0

C

+

3) 1

C

₂ 2

V

0

C

C

+

4) 0 1 2 1

_V

C

C

C

+

80. The capacitance of a parallel plate capacitor is 16µF. When a glass slab is placed between the plates, the potential difference reduces to 1/8th _{of the original value. What}

is dielectric constant of glass ?

1) 4 2) 8 3) 16 4) 32

81. The equivalent capacity across M and N in the given figure is:

1) 5C/3 2) 2/3C 3) C 4) 3/2C

82. Three plates of common surface area A are connected as shown. The effective capacitance between points P and Q will be:

1)

d

A

0

ε

₂₎

d

A

3

ε

₀ 3)

d

A

2

3

ε

₀ 4)

d

A

2

ε

₀

83. A dielectric slab of dielectric constant K = 5 is covered from all sides with a metallic foil. This system is introduced into the space of a parallel plate capacitor of capacitance 10 µF. The slab fills almost the entire space between the plates, but does not touch the plates. The capacitance will become nearly:

1) ∞ 2) zero 3) 2 pF 4) 50 Pf

84. A capacitor of capacitance 1 µF can withstand the maximum voltage 6 kV while a capacitor of capacitance 2.0 µF can withstand the maximum voltage 4 kV. If the two capacitors are connected in series, then the two capacitors combined can take up a maximum voltage of:

1) 2.4 kV 2) 5 kV 3) 9 kV 4) 10 kV

85. A spherical conductor of radius 2m is charged to a potential of 120 V. It is now placed inside another hollow spherical conductor of radius 6 m. Calculate the potential of bigger sphere if the smaller sphere is made to touch the bigger sphere:

1) 20 V 2) 60 V 3) 80 V 4) 40 V

1

IIT- PHY - CP

86. A parallel plate capacitor has two layers of dielectrics as shown in figure. Then the ratio of potential difference across the dielectric layers when connected to the battery is:

1) 2 1

K

K

2)

b

K

a

K

2 1 3)

a

K

b

K

1 2 ₄₎

b

K

a

K

1 2

LEVEL - III :

1. A parallel plate capacitor C is connected to a battery and it is charged to a potential difference of V. Another capacitor of capacitance 2C is similarly charged to a potential difference of 2V. The charging battery is now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to positive of the other. The final energy of the configuration is

A) zero B) 3/2 CV2 _{C) 25/6 CV}2 _{D) 9/2 CV}2

2. The amount of heat liberated when a capacitor of C farads charged to a potential difference of V volts is discharged through a resistor of R ohms is H joules. The same capacitor is now charged to a potential difference of 2V and discharged through a resistor of 2 R ohms, then heat liberated is

A) 4H B) 2H C) H D) H/2

3. A capacitor of capacity C1 is charged to a potential V0. The electrostatic energy

stored in it is U0. It is connected to another uncharged capacitor of capacitance C2

in parallel. The energy dissipated in the process is

A) 0 2 1 2

_U

C

C

C

+

B)

C

₁ 1

C

₂

U

0

C

+

C) 0 2 2 1 2 1 _U C C C C     + − _D) 0 2 1 2 1

_U

)

C

C

(

2

C

C

+

4. In the circuit shown in figure, if the key is turned so that instead of 1, 2 terminals 1, 3 are connected, then the heat liberated in resistor R is

A) C( ₁~ ₂)2 2 1 _{ε ε} B) C( _{1 2})2 2 1 _{ε +ε} C) ₁2 C ₂2 2 1 C 2 1 _{ε − ε} D) ₁2 C ₂2 2 1 C 2 1 _{ε + ε}

5. Three capacitors C1, C2, C3 are connected as

shown in figure to one another and to points P1, P2, P3 at potentials V1, V2 and V3. If the

IIT- PHY - CP

potential of the point O is A) 3 2 1 3 3 2 2 1 1

C

C

C

V

C

V

C

V

C

+

+

+

+

B) V1 + V2 + V3 C) 0 D) 2 1 3 1 3 2 3 2 1

C

_C

V

C

_C

V

_C

C

V

+

+

6. A parallel plate capacitor having capacitance C has two plates of same area A and thickness t. Figure shows the charges available on the four surfaces of the plates. The potential difference V between the two plates is given by

A)













−

C

q

q

2

1

2 3 _B)

_











−

C

q

q

_{2 3} C)













−

C

q

q

2

1

_{2 4} D)













−

C

q

q

_{2 4}

7. The plates of a parallel plate capacitor are separated by d cm. A plate of thickness

t cm with dielectric constant k1 is inserted and the remaining space is filled with a

plate of dielectric constant k2. If Q is the charge on the capacitor and area of plates

is A cm2 _{each, then potential difference between the plates is}

A)     ₊ − ε0 1 k2 t d k t A Q B)     ₊ − π 2 1 k t d k t A Q 4 C)     − + π t d k t k A Q 4 _{1 2} D)     ₊ − ε 2 1 0 k t d t k A Q

8. Find the capacitance of a system of three parallel plates each of area A separated by distances d1 and d2. The space between them is filled with dielectrics of relative

dielectric constants ε1 and ε2. The dielectric constant of free space is ε0

A) 1 2 2 1 0 2 1

d

d

A

ε

+

ε

ε

ε

ε

B) 2 2 1 1 0 2 1

d

d

A

ε

+

ε

ε

ε

ε

C) 2 1 2 1 0 2 1

d

d

)

(

A

ε

+

ε

ε

ε

ε

D) ) d d ( A 2 2 1 1 0 2 1ε ε ε +ε ε

9. Two identical capacitors, have the same capacitance C, One of them is charged to potential V1 and the other to V2. The negative ends of the capacitors are connected

together. When the positive ends are also connected, the decrease in energy of the combined system is A) C

(

V₁2 V₂2

)

4 1 _{− B)}

(

2

)

2 2 1 V V C 4 1 _{+ C)}

₍

₎

2 2 1 V V C 4 1 _{− D)}

₍

₎

2 2 1 V V C 4 1 ₊

10. One plate of a capacitor is connected to a spring as shown in the figure. Area of both the plates is A. In steady state, separation between the plates is 0.8 d (spring was unstretched and the distance between the plates was d when the capacitor was uncharged). The force constant of the spring is approximately.